A plano convex lens fits exactly in to a plano concave lens. Their plane surfaces are parallel to each other. Lenses are made up of different materials of refractive indices ' n 1 ' n 2

Question

A plano convex lens fits exactly in to a plano concave lens. Their plane surfaces are parallel to each other. Lenses are made up of different materials of refractive indices ' n 1 ​ ' n 2 ​ ' and ' R ' is the radius of curvature of the curved surface of lenses. Focal length of the combination is

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For the plano-convex lens, the focal length $f_{1}$ is given by the lens maker's formula:

$\frac{1}{f _{1}} = (n_{1} - 1) (\frac{1}{R} - \frac{1}{\infty}) = \frac{n _{1} - 1}{R}$

For the plano-concave lens, since it fits exactly into the plano-convex lens, its curved surface has the same radius of curvature $R$ but is concave in nature. Its focal length $f_{2}$ is given by:

$\frac{1}{f _{2}} = (n_{2} - 1) (- \frac{1}{R} - \frac{1}{\infty}) = - \frac{n _{2} - 1}{R}$

When the lenses are placed in contact, the equivalent focal length $F$ of the combination is:

$\frac{1}{F} = \frac{1}{f _{1}} + \frac{1}{f _{2}}$

$\frac{1}{F} = \frac{n _{1} - 1}{R} - \frac{n _{2} - 1}{R}$

$\frac{1}{F} = \frac{n _{1} - 1 - n _{2} + 1}{R} = \frac{n _{1} - n _{2}}{R}$

$F = \frac{R}{n _{1} - n _{2}}$

Answer: $\frac{R}{n _{1} - n _{2}}$

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